Power On Delay Circuit

[MedicMan]

Hi experts,

I have a simple circuit that enables a voltage regulator with a push button.
I want to be able to hold the same push button for a while for a hard reset in case the microcontroller downstream locks up.

The attached schematic turns the Vreg on when the button is pressed (a microcontroller holds the enable high after release) and if the button is held down for 10 secs forces it off.

The bit I need help with is that it only works correctly the first time when the cap is discharged. If I want to turn the circuit on again after a hard reset the time delay is very short as the cap is already charged...

Any thoughts on how to modify this circuit or a better idea greatly appreciated.

 

[barry]

You want to turn the regulator off if the button is held for greater than 10 seconds? But you say "Vreg on when the button is pressed". So, if I understand correctly, you've got the output of that diode connected to the output of your micro (hmmm....) and a momentary press of the button turns the regulator on , but a long press turns it off. Is that it?

 

[shegzzyy]

It wil only work well the first time.when the power is off,the capacitor isn't rapidly discharged enough to enable you obtain the required delay when you switch it on again.to solve this you need to place a diode between the base of the transistor and vcc.when the suply is present the diode doesn't conduct thus the capacitor charges up.when the supply is absent,the capacitor is discharged rapidly through the diode just like im my delay-on circuit below.hope it helps solve your problem

 

[MedicMan]

Hey shegzzyy,

Thanks for the suggestion. I did simulate it with the added Diode and found the capacitor is still not discharging. In the attached image you can see the initial delay time is around 9secs, but after release and press again the delay is 5secs.
I guess I dont understand how the diode would discharge the cap as it's connected to +ve. When the cap is fully charged it would also be at +ve so the diode path provides no potential difference???

 

[shegzzyy]

The circuit works perfectly.i too had a problem someyears ago trying to make a RC delay timer.when the suply voltage is at 0v{switch off},the base of the xsistor will be near vcc thus foward biasing the diode thus discharging the capacitor since curent always flow from high to low potential.which simulator do you use?

 

[MedicMan]

Hey again,

I'm using National Instruments Multisim. What I see happening: C1 is charged to VBatt while the switch is pressed. When the switch is released C1 discharges through Q1 base (takes a long time). D2 provides a forward bias from C1 to VBatt and hence does not aid faster discharge. The only way I see this circuit working correctly is if releasing the switch discharges C1 to ground quickly.... Did I understand the circuit you showed correctly?

I thought it might be something particular about my circuit so simulated your example exactly, but got the same results... The second time the switch is pressed the delay is greatly reduced....